Post on 07-Apr-2018
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WEEK 1
ELECTRIC CHARGE AND
ELECTRIC FIELD
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Lightning is a powerful natural electrostatic discharge produced during a thunderstorm.
INTRODUCTION
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Pr ope r ties of the elect r ic
cha rg eProton (positivecharge )
Charge value = 1.6 x 10 -19 CElectron (negativecharge )
Charge value = -1.6 x 10 -19 C
Total chargeNumber of charges
Charge for 1 proton/electron
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Lets t ry..For each of these pair of charges, select whether the force is attractive or repulsive.
+ -
+-+ +
- -
attraction
repulsion
attraction
repulsion
What we conclude about the type of charge between li ke charges and unli ke charges?Answer
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Lets t ry..For each of these pair of charges, select whether the force is attractive or repulsive.
+ -
+-+ +
- -
attraction
repulsion
attraction
repulsion
Answer : The force between li ke charges is repulsive and the force between unli ke charges isattractive.
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Coulomb s LawExperiment shows that the electric force betweentwo charges is proportional to the product of thecharges and inversely proportional to the squaredistance between them.
2
21
r
QQk F !
o
k TQ4
1! k is proportionality constant = 9.0 x 10 9 N.m 2 / C2
Yes! We can calculate the forces value by using
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Ex e r cise 1
Tw o isolated small objects have charges of 0.04 QC and 0.06 QC are 5cm apart. What is the vector componentof electrostatic force acting on each object?
r = 0.05 m
Q1 = + 0.04 C Q2 = -0.06 C
Solution
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Step 3
Write the vector component
F12Q1 F12 = 0.00864 N
the force direction isto + x
Hence, the vectorcomponent:- F12 = 0.00864
iStep 4
Calculate Force on Q 2
Q2 = -0.06 CF21
IF12I = IF21IF21 = -0.00864 i
i
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Ex e r cise 2
- + +
Calculate the resultant force on the charge Q 3 due toother t w o charges located as sho w n in figure above.Identify the direction of resultant/ net force on Q 3 .
Q1 = -4.2 C Q2 = + 1.3 C Q3 = + 1.1 C
1.0 cm2.0 cm
Solution
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Step 3
Write the vector component
Step 4
F31 F3
-103.95 i 1 8. 7 i
Total up all the forces
Fnet = F 31 + F3= (-103.95 i) + ( 1 8. 7 i)= 4.75 i
Hence, the M AG NITU D E of total Force = 4.75 N
the D IR ECT ION of total Force= + x direction
+
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Three point charges arearranged as sho w n inFigure belo w . Find the
magnitude and directionof the net force on chargeQ A
+
+-
4 m
3 m
Q A
=+
12 C
Q C=+ 20 CQ B=-16 C
Ex e r cise 3
Solution
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Step 1D raw the direction of the electric force on Q A due to Q B and Q C
4 m
3 m
Q A=+ 12 C
Q C=+ 20 CQ B=-16 C
FAC
+
+-
FA B
Click on charge Q C and Q Bto observe the direction of force experience by chargeQA
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Q A=+ 12 C
FAC
+
FAB
Step 2
Force Calculati on Meth od Value
FAC
FAB
2k
A C Q Q
r
2k
A BQ Qr
9 6 6
2
9 0 ( 2 1 0 )(2 0 1 0 )4
v v v
9 6 6
29 10 (12 10 )(16 10 )
5
v v v
Find the magnitude of force onQ A due to QB and QC
0.135 N
0.069 N
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Step 3Write the vector component
Q A
FAC = 0.135 N
+
FAB = 0.069 N
Magnitu de X - compo nent Y - compo nent
FAC= 0.135 N
FAB = 0.069 N
Total up
53.13 o
0.135 cos 90 o= 0 i 0.135 sin 90 o = 0.135 j
0.069 cos 233.13 o= -0.04 i 0.069 sin 233.13 o = - 0.055 j
= - 0.04 i = 0.08 j
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Step 4
Find the magnitude from the vector component calculated
Hence,
2 2(0.04) (0.08)
Vector component = - 0.04 i + 0.08 j
Magnitude =
= 0.089 N
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E lect r ic field lines
Positive charge Negative charge
Increase charge
An electric field is a region where a charge experience a force.
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E lect r ic field lines
+
Positive charge Negative charge
What is the effect of the increasing the charge of the field lines?Answer
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E lect r ic field lines
-
Positive charge Negative charge
Increase charge
What is the difference between the electric field due to positive and negative charge?Answer
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E lect r ic field lines
-
Positive charge Negative charge
Increase charge
Answer: The directions of the field lines are opposite. Electric field experienced by thepositive charge is radially outward while electric field experienced by the negative charge isradially inward.
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E lect r ic field lines
-
Positive charge Negative charge
Answer: The directions of the field lines are opposite. Electric field experienced by thepositive charge is radially outward while electric field experienced by the negative charge isradially inward.
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E lect r ic field
Direction of electric field around the charge Q can be visualized by placing smapositive charge into the field and displaying the force vector acting on each.
-
Positive charge
Negative chargeQ
A
B
C
+
+
+
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E lect r ic field st r en g thElectric field strength Eat a point is defined asthe force acting on aunit positive charge atthat point.
+ +
r
2
21
r
QQk !
Q1 Q2
Derivation
Force on test charge Q 2 at a distance rfrom a point charge Q 1 is
2
21
r
QQk !
Electric field strength
2
2
21
2
1
Qr Qk Q
Q F
E !!
2
1
r
Qk E !Hence,
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2r kQ
E p !
Step 1D raw the direction of the electric field on P due to Q
P _Q = -3.0 x 10 -6 C
Imagine there is a + vecharge here
+EP
StepCalculate the magnitude of the electric field on P
P+
EP +
9 x 10 9 Nm 2 / C2 Q = -3.0 x 10 -6 C
0.3 m
Ep = 3.0 x 105 N/C
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Step 3
Write the vector component
PEP ++
The irectionof E is to -X Ep = 3. x 1
5
/E p = -3. x 1 5 i /Hence, the vector
component isF inal Answer
Magnitude = 3.0 x 105
N/C anddirection = to X axis (to the left)
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S ummary The force between li ke charges is repulsive and the forcebetween unli ke charges is attractive.Coulomb s law states that the electric force between twocharges is proportional to the product of the charges and
inversely proportional to the square distance betweenthem.
Electric field strength E at a point is defined as the forceacting on a unit positive charge at that point.
2
21
r
QQk !
2r
Qk
q F
E !!
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S ummary
Electric field is a vector quantity. Its directionis the direction of force on a positive charge.
Unit for electric field is N/C or V/m.